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A thin uniform rod of mass \(M\) and length \(L\) is bent at its center so that the two segments are now perpendicular to each other. Find its moment of inertia about an axis perpendicular to its plane and passing through (a) the point where the two segments meet and (b) the midpoint of the line connecting its two ends.

Short Answer

Expert verified
(a) \(\frac{1}{3}ML^2\) (b) \(\frac{5}{48}ML^2\)

Step by step solution

01

Understanding the Problem

We need to find the moment of inertia of a bent rod that forms an 'L' shape. This involves considering the moments of inertia for each segment of the rod about different axes.
02

Applying Perpendicular Axis Theorem

Since the rod is in a plane, we apply the perpendicular axis theorem. It states that for thin, flat objects in the plane, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia about two perpendicular axes in the plane that intersect at the point where the perpendicular axis goes through.
03

Calculating Moment of Inertia About the Meeting Point

For the first part, where the perpendicular axis passes through the point where the segments meet, consider each segment separately. Each segment, a rod of length \(L/2\), has a moment of inertia \(I = \frac{1}{3}ml^2\) about an axis through one end. Thus, for each segment, \(I = \frac{1}{3} \frac{M}{2} \left( \frac{L}{2} \right)^2\). The total moment is the sum for both segments.
04

Moment of Inertia Through the Meeting Point Calculation

Calculate the individual moments: \(I_A = \frac{1}{3} \cdot \frac{M}{2} \cdot \left( \frac{L}{2} \right)^2\) for each segment. Since there are two segments, double this result to find \[ I_{total} = 2 \cdot \frac{1}{6}ML^2 = \frac{1}{3}ML^2. \]
05

Calculating Moment of Inertia About the Midpoint of Connection Line

For the axis through the midpoint of the line connecting the two endpoints, use parallel axis theorem to shift the calculated moments from one end to the midpoint. For a rod, shifting an end axis to another point involves \(I' = I_{cm} + md^2\), where \(d\) is the distance to the new axis.
06

Shifting Axes to the New Midpoint

For each segment, its center of mass is at \(L/4\) from the corner. Shift from the meeting point: \[ I_{new} = \frac{M}{2} \left( \frac{L}{4} \right)^2.\]Use symmetry to find this distance aids in halving again.Therefore the sum of moments is \[ I_{midpoint} = 2 \cdot \left( \frac{1}{3} \frac{M}{2} \left( \frac{L}{2} \right)^2 + \frac{M}{2} \left( \frac{L}{4} \right)^2 \right). \]
07

Calculating Total Inertia for Midpoint

Substitute and calculate: \[ I_{midpoint} = 2 \cdot \left( \frac{1}{6}ML^2 + \frac{1}{32}ML^2 \right) = 2 \cdot \frac{5}{96}ML^2 = \frac{5}{48}ML^2. \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Perpendicular Axis Theorem
The perpendicular axis theorem is a valuable tool in the study of rotational motion for planar objects. It states that the moment of inertia of a flat object about an axis perpendicular to its plane is equal to the sum of the moments of inertia about two perpendicular axes lying in the plane of the object and intersecting at the point where the perpendicular axis passes through.
For example, in the case of a thin uniform rod, when bent into an L-shape, this theorem helps in breaking down the problem into simpler parts. The rod segments are perpendicular to each other, aligned with axes in the plane. Applying the theorem allows calculation of moments of inertia by summing the contributions from these in-plane axes.
  • This theorem is particularly useful when dealing with composite shapes.
  • It simplifies calculations by allowing us to consider each segment separately.
  • Thus, it transforms a complex problem involving a seemingly irregular shape into an easily manageable one.
Parallel Axis Theorem
The parallel axis theorem is another vital principle used to find the moment of inertia when the axis of rotation is shifted parallel to an axis that passes through the center of mass. It states: \[I' = I_{cm} + md^2\] where \(I'\) is the moment of inertia about the new axis, \(I_{cm}\) is the moment of inertia through the center of mass, \(m\) is the mass of the object, and \(d\) is the perpendicular distance between the two axes.
When calculating the moment of inertia for an L-shaped rod about a point other than the meeting point of the segments, the parallel axis theorem becomes essential. Each segment's inertia needs adjustment to account for the lateral ``shift'' in the axis.
  • This shift often arises when calculating for axes through points like the midpoint of the line connecting the device ends.
  • The theorem lets us maintain consistency in calculations by accurately adjusting the inertia as needed.
L-Shaped Rod
An L-shaped rod is formed when a uniform thin rod is bent at its center to create two perpendicular arms. This configuration represents a simple composite object whose inertia can be quite complex to derive.
An L-shaped rod shares characteristics of both linear and angular structures. Each leg or segment has attributes applicable to straight rods, but when combined in an L-shape, the calculation of overall inertia requires consideration of both geometrical arrangement and properties of individual segments.
The L-shape necessitates the use of both the perpendicular and parallel axis theorems. Each segment contributes differently depending on its orientation and the selected axis.
  • A segment can be visualized as a simple rod, but its placement or positioning in the L-configuration brings unique contributions.
  • Analyzing segment placement helps in applying the axis theorems effectively, leading to precise calculations.
Calculation of Segments
In calculating the moment of inertia for an L-shaped rod, it's essential to consider each segment individually. Each segment is essentially a rod which can be treated separately from the other.
Calculations generally start with deriving the moment of inertia of each segment about its axis passing through one end, then summing them up to get the total inertia of the combined shape.
  • Initially, determine the inertia of each segment's center of mass.
  • Then apply the parallel axis theorem to shift the axis as required by the problem.
  • This method simplifies complex shapes by treating them as a sum of simpler parts, thus enabling effective and accurate inertia calculations.
In our exercise, this involves computing the moment of inertia for each of the perpendicular segments of the L-shaped rod separately and then considering their contribution to the total system.

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Most popular questions from this chapter

The flywheel of a gasoline engine is required to give up 500 \(\mathrm{J}\) of kinetic energy while its angular velocity decreases from 650 \(\mathrm{rev} / \mathrm{min}\) to 520 \(\mathrm{rev} / \mathrm{min}\) . What moment of inertia is required?

A disk of radius 25.0 \(\mathrm{cm}\) cm is free to turn about an axle perpendicular to it through its center. It has very thin but strong string wrapped around its rim, and the string is attached to a ball that is pulled tangentially away from the rim of the disk (Fig. P9.65). The pull increases in magnitude and produces an acceleration of the ball that obeys the equation \(a(t)=A t,\) where \(t\) is in seconds and \(A\) is a constant. The cylinder starts from rest, and at the end of the third second, the ball's acceleration is 1.80 \(\mathrm{m} / \mathrm{s}^{2}\) . (a) Find \(A .\) (b) Express the angular acceleration of the disk as a function of time. (c) How much time after the disk has begun to turn does it reach an angular speed of 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (d) Through what angle has the disk turned just as it reaches 15.0 \(\mathrm{rad} / \mathrm{s} ?\) (Hint: See Section \(2.6 .\) .

A wheel changes its angular velocity with a constant angular acceleration while rotating about a fixed axis through its center. (a) Show that the change in the magnitude of the radial acceleration during any time interval of a point on the wheel is twice the product of the angular acceleration, the angular displacement, and the perpendicular distance of the point from the axis. (b) The radial acceleration of a point on the wheel that is 0.250 \(\mathrm{m}\) from the axis changes from 25.0 \(\mathrm{m} / \mathrm{s}^{2}\) to 85.0 \(\mathrm{m} / \mathrm{s}^{2}\) as the wheel rotates through 20.0 rad. Calculate the tangential acceleration of this point. (c) Show that the change in the wheel's kinetic energy during any time interval is the product of the moment of inertia about the axis, the angular acceleration, and the angular displacement. (d) During the 20.0 -rad angular displacement of part (b), the kinetic energy of the wheel increases from 20.0 \(\mathrm{J}\) to 45.0 \(\mathrm{J} .\) What is the moment of inertia of the wheel about the rotation axis?

A roller in a printing press turns through an angle \(\theta(t)\) given by \(\theta(t)=y t^{2}-\beta t^{3},\) where \(\gamma=3.20 \mathrm{rad} / \mathrm{s}^{2}\) and \(\beta=0.500 \mathrm{rad} / \mathrm{s}^{3}\) (a) Calculate the angular velocity of the roller as a function of time. (b) Calculate the angular acceleration of the roller as a function of time. (c) What is the maximum positive angular velocity, and at what value of \(t\) does it occur?

A thin, rectangular sheet of metal has mass \(M\) and sides of length \(a\) and \(b .\) Use the parallel-axis theorem to calculate the moment of inertia of the sheet for an axis that is perpendicular to the plane of the sheet and that passes through one corner of the sheet.

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